Math Problem Statement

Solve

An experiment yielded n = 12 temperature measurements, assumed to have a normal distribution with known standard deviation σ = 6.88. Calculate: (a) The 99% two-sided confidence interval on the mean temperature. (b) The 95% confidence single-sided lower bound on the mean. (c) For 95% confidence that the estimated mean error E < 1, determine the required sample size.

Solution

To solve this problem, let’s go through each part step by step.

Step 1: Find the Mean of the Sample Data

The temperatures measured are: 77.5,74.9,67.3,83.9,72.2,86.7,80.4,71,72.9,86.7,73.6,86.877.5, 74.9, 67.3, 83.9, 72.2, 86.7, 80.4, 71, 72.9, 86.7, 73.6, 86.8

Let's calculate the sample mean xˉ\bar{x}: xˉ=77.5+74.9+67.3+83.9+72.2+86.7+80.4+71+72.9+86.7+73.6+86.812\bar{x} = \frac{77.5 + 74.9 + 67.3 + 83.9 + 72.2 + 86.7 + 80.4 + 71 + 72.9 + 86.7 + 73.6 + 86.8}{12}

Step 2: Construct a 99% Confidence Interval for the Mean

  1. The sample size n=12n = 12.
  2. The population standard deviation σ=6.88\sigma = 6.88.
  3. For a 99% confidence level, the z-score (critical value) zα/2z_{\alpha/2} is approximately 2.576.

The formula for a confidence interval is: xˉ±zα/2σn\bar{x} \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}

Step 3: Calculate the 95% Single-Sided Lower Bound

For a single-sided lower bound with 95% confidence, we use a z-score of approximately 1.645.

The lower bound is: xˉzσn\bar{x} - z \frac{\sigma}{\sqrt{n}}

Step 4: Determine the Sample Size Needed for a Desired Margin of Error

We need to find the required sample size nn to have a margin of error E<1E < 1 with 95% confidence. The formula for sample size with a given margin of error is: n=(zσE)2n = \left( \frac{z \sigma}{E} \right)^2 where E=1E = 1 and z=1.645z = 1.645.

Let me calculate each of these steps.Here are the results for each part:

  1. Sample Mean (xˉ\bar{x}): 77.825

  2. 99% Confidence Interval for the Mean: 72.71μ82.9472.71 \leq \mu \leq 82.94

  3. 95% Single-Sided Lower Bound for the Mean: μ74.56\mu \geq 74.56

  4. Required Sample Size for Margin of Error E<1E < 1 with 95% Confidence: n=129n = 129

If you need more details or have questions, let me know!

Related Questions

  1. What does a 99% confidence interval indicate about the mean?
  2. How is the margin of error affected by the sample size?
  3. Why is a larger sample size required for a smaller margin of error?
  4. How would the confidence interval change if the standard deviation were different?
  5. What if we wanted a 90% confidence interval—how would the z-score change?

Tip

Remember, confidence intervals provide a range within which we expect the true population parameter to lie, based on the sample data and a specified confidence level.

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Math Problem Analysis

Mathematical Concepts

Statistics
Confidence Intervals
Sample Mean
Margin of Error

Formulas

Confidence Interval formula: \( \bar{x} \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}} \)
Lower Bound formula: \( \bar{x} - z \frac{\sigma}{\sqrt{n}} \)
Sample Size for Margin of Error: \( n = \left( \frac{z \sigma}{E} \right)^2 \)

Theorems

Central Limit Theorem
Properties of Normal Distribution

Suitable Grade Level

Undergraduate Statistics